Question

Integral of 11 ln(cube root(x)) dx?

Answer 1

$\bf \int [11\ ln\left( \sqrt[3]{x} \right)]dx\qquad \begin{cases} ln\left( \sqrt[3]{x} \right)\implies ln\left( x^{\frac{1}{3}} \right) \\ \quad \\ \frac{1}{3}ln(x) \end{cases}\qquad thus \\ \quad \\\\ \quad \\ \int [11\ ln\left( \sqrt[3]{x} \right)]dx\implies \int[11\cdot \frac{1}{3}ln(x)]dx\impliedby \textit{scalars to the left} \\ \quad \\ 11\cdot \frac{1}{3}\int[ln(x)]dx\implies \cfrac{11}{3}\int[ln(x)]dx$

and I'm pretty sure you know what that is

Answer 2

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int 11\,\ell n(^3\hspace{-5}\sqrt{x})\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(^3\hspace{-5}\sqrt{x})\cdot 11\,dx\qquad\quad(i)}$


Integration by parts:

$\begin{array}{lcl} \mathsf{u=\,\ell n(^3\hspace{-5}\sqrt{x})}&\quad\Rightarrow\quad&\mathsf{du=\dfrac{1}{^3\hspace{-5}\sqrt{x}}\cdot \dfrac{d}{dx}(^3\hspace{-5}\sqrt{x})\,dx}\\\\ &&\mathsf{du=x^{-1/3}\cdot \dfrac{1}{3}\,x^{(1/3)-1}\,dx}\\\\ &&\mathsf{du=\dfrac{1}{3}\,x^{-1/3}\cdot x^{-2/3}\,dx}\\\\ &&\mathsf{du=\dfrac{1}{3}\,x^{-1}\,dx} \\\\\\ \mathsf{dv=11\,dx}&\quad\Leftarrow\quad&\mathsf{v=11x} \end{array}$


$\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\ell n(^3\hspace{-5}\sqrt{x})\cdot 11\,dx=\ell n(^3\hspace{-5}\sqrt{x})\cdot 11x-\int\!11x\cdot \frac{1}{3}\,x^{-1}\,dx}\\\\\\ \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!11\,\ell n(^3\hspace{-5}\sqrt{x})\,dx=11x\,\ell n(^3\hspace{-5}\sqrt{x})-\frac{11}{3}x+C}\end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}$


I hope this helps. =)